To solve this problem, we need to split the integrals and evaluate each one using the Beta function.

Problem Statement

Given integrals:

1. $\int_{0}^{3} \frac{x^{3/2}}{\sqrt{3 - x}} \, dx$
2. $\int_{0}^{1} \frac{1}{\sqrt{1 - x^{1/4}}} \, dx$

The Beta function can be represented as: $B(x, y) = \int_{0}^{1} t^{x-1} (1 - t)^{y-1} \, dt$ We will try to manipulate each integral into this form.

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1. First Integral

$\int_{0}^{3} \frac{x^{3/2}}{\sqrt{3 - x}} \, dx$

To use the Beta function, we can substitute $x = 3u$, so that $dx = 3 \, du$ and the limits change from $x = 0$ to $x = 3$, which gives $u = 0$ to $u = 1$. Substituting these, we get: $\int_{0}^{1} \frac{(3u)^{3/2}}{\sqrt{3 - 3u}} \cdot 3 \, du$ Simplifying: $= 3^{5/2} \int_{0}^{1} u^{3/2} (1 - u)^{-1/2} \, du$ This is now in the form of a Beta function: $3^{5/2} \cdot B\left( \frac{5}{2}, \frac{1}{2} \right)$

Using the property of the Beta function in terms of Gamma functions, we have: $B\left( \frac{5}{2}, \frac{1}{2} \right) = \frac{\Gamma\left( \frac{5}{2} \right) \Gamma\left( \frac{1}{2} \right)}{\Gamma\left( 3 \right)}$

Now, we calculate:

• $\Gamma\left( \frac{5}{2} \right) = \frac{3\sqrt{\pi}}{4}$
• $\Gamma\left( \frac{1}{2} \right) = \sqrt{\pi}$
• $\Gamma(3) = 2$

Thus: $B\left( \frac{5}{2}, \frac{1}{2} \right) = \frac{\frac{3\sqrt{\pi}}{4} \cdot \sqrt{\pi}}{2} = \frac{3\pi}{8}$

Therefore, the first integral becomes: $3^{5/2} \cdot \frac{3\pi}{8} = \frac{243\pi}{8}$

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2. Second Integral

$\int_{0}^{1} \frac{1}{\sqrt{1 - x^{1/4}}} \, dx$

Let $x = t^4$, so $dx = 4t^3 \, dt$. The limits change as follows:

• When $x = 0$, $t = 0$
• When $x = 1$, $t = 1$

Substituting into the integral, we get: $\int_{0}^{1} \frac{1}{\sqrt{1 - t}} \cdot 4t^3 \, dt = 4 \int_{0}^{1} t^3 (1 - t)^{-1/2} \, dt$ This is now in the form of a Beta function: $4 \cdot B\left( 4, \frac{1}{2} \right)$

Using the Beta function in terms of Gamma functions, we have: $B\left( 4, \frac{1}{2} \right) = \frac{\Gamma(4) \Gamma\left( \frac{1}{2} \right)}{\Gamma\left( \frac{9}{2} \right)}$

Now, we calculate:

• $\Gamma(4) = 6$
• $\Gamma\left( \frac{1}{2} \right) = \sqrt{\pi}$
• $\Gamma\left( \frac{9}{2} \right) = \frac{105 \sqrt{\pi}}{16}$

Thus: $B\left( 4, \frac{1}{2} \right) = \frac{6 \cdot \sqrt{\pi}}{\frac{105 \sqrt{\pi}}{16}} = \frac{96}{105} = \frac{32}{35}$

Therefore, the second integral becomes: $4 \cdot \frac{32}{35} = \frac{128}{35}$

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Final Answer

The value of the two integrals are:

1. $\int_{0}^{3} \frac{x^{3/2}}{\sqrt{3 - x}} \, dx = \frac{243\pi}{8}$
2. $\int_{0}^{1} \frac{1}{\sqrt{1 - x^{1/4}}} \, dx = \frac{128}{35}$

Would you like further clarification on any part? Here are some follow-up questions to extend your understanding:

1. How does the Beta function relate to the Gamma function in integration problems?
2. Can we solve similar integrals using the Beta function directly, or are substitutions always required?
3. What are some practical applications of the Beta and Gamma functions in probability and statistics?
4. How do properties of the Beta function simplify integrals with certain forms?
5. How would changing the limits of integration affect the steps to solve these integrals?

Tip: When tackling integrals that resemble the Beta or Gamma function form, look for substitutions to simplify them into known Beta function formats.